help-10

[robust locust]

yea

[fallow reef]

What did you set it equal to?

[robust locust]

some quadratic equation

i got 54=x^2+3x

so it turns to x^2+3x+54

[fallow reef]

... where did 54 come from?

AB * AC = 45

[robust locust]

oh

OPH

[red ice]

Nah you're tripping, 3(3 + 15) = 54

[robust locust]

what

[fallow reef]

um...

[robust locust]

54 is AB*AC

18x3

[red ice]

Ah hold on

[fallow reef]

ah dammit, you're right

sorry

[red ice]

Nah it's not 3 * 15 guys

[high lily]

you didn't manipulate that correctly

[red ice]

Yeah the issue here is you need to subtract by 54

0 = x^2 + 3x - 54

[robust locust]

yea sorry

i mistyped

but i did subtract it

[red ice]

Then you shouldn't have x^2 + 3x + 54 = 0

[robust locust]

yep

i mistyped the 54

[fallow reef]

So.. you're on the right track -

1. What was your positive result?
2. Why did you think your positive result was wrong?

[robust locust]

6

[robust locust]

it says it on the thing

[high lily]

what thing

recall what x represents and what the question is asking for

[fallow reef]

I haven't seen anything except the original problem, without any problems worked out

[robust locust]

oh

[robust locust]

thanks

that was the issue

this is what it looks like

.close

[obtuse pebble]

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[obtuse pebble]

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[halcyon jewel]

help

[halcyon jewel]

step by step w out giving me the answer,

ngl just where do i start

[nocturne schooner]

take derivitive

-4/13.5

[halcyon jewel]

yea im not gonna waste peoples time

.close

[obtuse pebble]

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[twilit meteor]

When a 3 digiit number is taken and reversed and this number is subtracted from the original number, and this number is then reversed again and added to the prev. number , u always get 1089. how do u prove ths?

[twilit meteor]

there are exceptions but it works for most numbers

for example take 694
reversed: 496
difference: 198
reversed: 891
sum:1089

[obtuse pebble]

@twilit meteor Has your question been resolved?

[naive prism]

this is a tough one, been thinking for a while

[obtuse pebble]

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[static furnace]

What was the final form of your equation before you found the variables k, h, and p

i found the same k and h, but a different p

[obtuse pebble]

@digital comet Has your question been resolved?

[frosty parcel]

it does, thanks a lot!, so basically just put it into e and fraction forms, then L'H. Cool!

this, on roblox

[obtuse pebble]

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[digital comet]

.reopen

[obtuse pebble]

✅

[digital comet]

What did u get for p

[obtuse pebble]

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[sturdy jasper]

I need help finding the explicit formula for a sequence

[sturdy jasper]

I'm working with this sequence, but I don't know how to begin finding the formula

I know that the negative signs are alternating, so there's likely a (-1)^(k-1)

My best guess is that the numerator should be 1, since its unchanging

I'm not really sure whats happening with the denominator

[thin stone]

look at each term as two different terms (left and right)

[sturdy jasper]

I see that they seem to be counting up in alternating fashion

[thin stone]

for example, the left part of the first term is 1/1, the second is 1/2, etc

[sturdy jasper]

1, 2, 3 ...
and
4, 5, 6 ...

[thin stone]

yes great

thats how you can construct the general term

[sturdy jasper]

I'm just not sure how to begin doing that

[thin stone]

okay sure

look at the difference in the denominators of each term

[sturdy jasper]

looks like a difference of 3, 2, 3, 2...

[thin stone]

isnt the difference always 3

[sturdy jasper]

I don't see that

i see that from 1 to 1/4, the denominator changed by 3, then from 4 to 2, it changes by 2, and repeats

[thin stone]

no no

the first term is 1 and 1/4 the second is 1/2 and 1/5

the third is 1/3 & 1/6

etc

[sturdy jasper]

oh i didnt notice the fact that two fractions are being subtracted within each term

so then the order would be (3/4), (3/10), (3/18), (3/28)...

[thin stone]

you dont have to subtract them, just look at how the denominators always have a difference of 3 :)

[sturdy jasper]

yeah, within each term
i dont know how i could translate that information into a formula without subtracting them though

[thin stone]

like for example i can rewrite the first term as (i'll call it $a_1$) $a_1 = \frac{1}{1} - \frac{1}{1+3}$

[warm shale]

anonymemes

[thin stone]

try to rewrite $a_2$

[warm shale]

anonymemes

[sturdy jasper]

like this?

[thin stone]

yes great

can you see the pattern now

[sturdy jasper]

a little bit

you can write it like this i assume?

[thin stone]

yes

[sturdy jasper]

and then making the negative sign alternate like this

actually thats not necessary

i just realized

it stays positive each term

[thin stone]

yeah this is your full answer :)

[sturdy jasper]

well thank you for helping me find it

:)

.close

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[jagged hawk]

how would you solve this?

[dense imp]

if the event "only one converts his goal" happened, this could take place two ways:
Jack scores and Ben doesn't OR Ben scores and Jack doesn't. how would you calculate the probabilities of these two using the info given in 1st sentence?

[jagged hawk]

0.3x0.6 + 0.7x0.4

what values do u sub in for P(A|B)=P(AB)/P(B) ?

[dense imp]

so the probability of one happening is 0.18 and the other 0.28, but we know this event occurred (one hit the ball) so the probability of it being Ben, (the 60% guy) would be simply: 0.18/(0.18+0.28)

[obtuse pebble]

@jagged hawk Has your question been resolved?

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[hard citrus]

can someone please help me with this question?

[hard citrus]

wait nvm

.close

[obtuse pebble]

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[primal idol]

Did i do it right

[high lily]

no

[primal idol]

How

[high lily]

what are you doing with the
6^(36x)
none if it seems valid

[primal idol]

Do i take the natural log of the whole thing then?

[surreal forge]

do u know this rule

[primal idol]

So

According to that rule

Is it just

Sorry guys i cant believe i cant solve it

Im taking upper core mechanical engineering (dynamics) classes and i find it weird that i dont use as much calculus for it

[high lily]

not quite,
you'd also need to multiply by derivative of 36x because chain rule

[obtuse pebble]

@primal idol Has your question been resolved?

[primal idol]

Im having a hard time

Understanding it

And ive taken Advanced Engineering Math

Sorry

[fallow reef]

How many nests do you have?

[primal idol]

Nests?

[fallow reef]

One function, nested in another function, nested in another function, etc.

your goal is to find the derivative of each "level" on its own, in series (in a chain.. hence, the chain rule)

d/dx f(g(h(x))) =

[primal idol]

Yeah

[fallow reef]

start on the outside.. f'(g(h(x))...

then go to the next one g'(h(x))..

h'(x)

[primal idol]

Im just lost

So this is wrong?

[fallow reef]

First part is

Left side needs to be addressed

[obtuse pebble]

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[rose ferry]

hey i am stuck what do I do next

[modern onyx]

if 3 is ur remainder

U would just write it as 3/x-1

[rose ferry]

why

soz i don’t know a lot about long division

so for any remainder we have to divide it by the whole like thing

[modern onyx]

yea

[rose ferry]

ah okay

[modern onyx]

Math

[rose ferry]

ur teaching me things my teacher didn’t

lmao

okay thanks

.close

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[coarse compass]

Hello! I need some university-lvl help I need to find the lateral surface area (I dont even know if I am using the right terms lol) of my eye's cornea that gets in contact with my soft contact lens. I found that the cornea is a prolate spheroid so I guess I can divide it to 2 to find it but I dont know if I should start by finding lateral surface area of the prolate spheroid. I hope you got what I am trying to say Any help is appreciated!!

[obtuse pebble]

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[coarse compass]

.close

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[cloud stream]

hi, im struggling to understand how distributive law was used here, i jsut cant seem to visualize how it was done. can anyone help D:

[obtuse pebble]

@cloud stream Has your question been resolved?

[obtuse pebble]

@cloud stream Has your question been resolved?

[cloud stream]

.close

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[grizzled tinsel]

if x^3 + px + q has a double root, show that the double root must be -3q/2p, where p cant be 0

[inland matrix]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[grizzled tinsel]

2

[inland matrix]

What have you tried so far?

[grizzled tinsel]

i made P'(x)

so that equal to 0

should be the double root

[inland matrix]

Doesn't double root mean 2 roots that are the same?

[grizzled tinsel]

yes

and i subbed

-3q/p into it but the equation is coming out too messy

so i dont think i should do that

[inland matrix]

That would be the root of P'(x), not p(x) though

If u sub the value of x you got then those will be points of maxima/minima

[grizzled tinsel]

what

if i put the derivative of (x^3+px+q)

then P(x) = P'(x) gives me x

[cyan void]

Well yeah, while you might be able to solve it that way

Might not be ideal

[grizzled tinsel]

yeah it def wasnt

is there another way

or a more proper way

[cyan void]

Notice that P'(x) is def simpler

[grizzled tinsel]

yeah

its 3x^2 + p

[cyan void]

The double root is both root of P(x) and P'(x)

[grizzled tinsel]

but

do i construct an equation or smth

[inland matrix]

Wait so it doesn't mean that P(x) has two roots which are same?

Mb then

[grizzled tinsel]

i alr made x^3 + px + q = 3x^2 + p

[cyan void]

It's an obvious consequence of being a double root.

[cyan void]

Hmm, wouldn't that be dead end as well

[grizzled tinsel]

yeah

[inland matrix]

I guess so

[grizzled tinsel]

im like

stuck

[inland matrix]

Assume the roots to be a,a and b then ig

[grizzled tinsel]

still confusing 😭

[marsh geyser]

Are u sure u copied correctly the question?

[grizzled tinsel]

yes
OH

ok

i show that

[cyan void]

Ah right. That is def a possibility

[grizzled tinsel]

the double root

is -3q/2p

not -3q/p

[surreal forge]

that makes more sense

[grizzled tinsel]

howd u figure that out wtf

are u psycic

psychic

[surreal forge]

when i did this problem i got -3q/2p as the root

[cyan void]

Ahh

[grizzled tinsel]

omg

sorry

[surreal forge]

anyway i think viete's relations provides fairly simple way of solvingthis

[marsh geyser]

Its ok

[grizzled tinsel]

is that the

[inland matrix]

.

[cyan void]

Hm, is that way simpler.

[marsh geyser]

Next time screenshot so i dont have to guess u are wrong

xD

[surreal forge]

loll

[grizzled tinsel]

a + b + y = -b/a

except the first is alpha and beta

[inland matrix]

Ye

[grizzled tinsel]

but i dont have bx^2

do i jsut let it 0

[inland matrix]

ye

[grizzled tinsel]

ok

then i do like

simulatneous or smth

[inland matrix]

Ye

[grizzled tinsel]

ight

ill try

should i sub the -3q/2p as a root

like

am i allowed to do that

if i have to show

[inland matrix]

That's the end result

Which you've to prove for it to be true

So no

[grizzled tinsel]

man

ok ty

[inland matrix]

Np

[grizzled tinsel]

done

ty

.close

[obtuse pebble]

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[grizzled tinsel]

The diagram shows the curve y=x^3 - x +3 and the point P(-2,-3) on the curve. The line L cuts the curve at P, and is tangent to the curve at another point A on the curve. Find the equation of the line L

[grizzled tinsel]

ignore the random bit below the graph they misplaced it on the wrong question (thats working out from it)

[harsh remnant]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[grizzled tinsel]

2

but basically 1 coz i just did 1 step

[harsh remnant]

How did you start off?

[grizzled tinsel]

just got derivative of x^3 - x +3

so i have

the gradient

but not the point

[harsh remnant]

We know the slope of the line

That should equal f'(A)

[harsh remnant]

Given in point P

[grizzled tinsel]

yes

but im confused as to like

how to get the other point

[harsh remnant]

What's the derivative?

[grizzled tinsel]

y' = 3x^2 -1

[harsh remnant]

What should that equal to

[grizzled tinsel]

that?

nots ure what to sub for x

just coz the point P

isnt the other intersection

that would give it

do i like

leave (3x^2-1) as a variable

as the gradient

then point gradient with -2,-3

[timid silo]

set the tangent point as (t,f(t))

L = f'(t)(x-t) + f(t)

put (-2,-3) in it

an solve

[grizzled tinsel]

ill try

[inland matrix]

You might get a cubic but you can use rrt in this case

[grizzled tinsel]

wahts rrt

[inland matrix]

Rational root theorem

[grizzled tinsel]

never heard of it

[grizzled tinsel]

do i put -2,-3 as like

L = -3

x = -2

[timid silo]

ye

[grizzled tinsel]

but then it comes out as

oh

hol don

but hten it comes out as lke

-3 = f'(t)(-2-t) + f(t)

[timid silo]

solve for t

[grizzled tinsel]

how can i find f'(t) if i dont know what t is

wait i can solve frm there?

[timid silo]

ye

[grizzled tinsel]

ok hold no

i dont know how to solve from there 😭

[inland matrix]

We have f'(t)

It's the same as y'

But replace x with t

[grizzled tinsel]

wait

like

y' as in 3x^2 -1

[inland matrix]

Yes

[grizzled tinsel]

OH

ok hold on

and i solve for x right

[inland matrix]

No

Since it's f'(t), replace x with t

Note that we assumed t to be the x coordinate of the point we wanna find

[grizzled tinsel]

eys

oh

like

all x for t

?

[inland matrix]

Yeah

[grizzled tinsel]

ok

[inland matrix]

3t²-1

[grizzled tinsel]

right

so

x = -2 and 1

do i find the gradient from here

coz

1 is my x coord for the other point

[inland matrix]

You.. didn't get a cubic?

In between?

[grizzled tinsel]

no i did get a cubic

was

2x^3 + 6x^2 - 8 = 0

[inland matrix]

Right

[inland matrix]

Yeah

[grizzled tinsel]

right

so im pre much done now

[inland matrix]

You can coz you have the x coordinate

[grizzled tinsel]

ok

great

[inland matrix]

How did u solve it then

I thought you never heard or rrt

[grizzled tinsel]

i used a calculator 😐

is rrt

like

helpful in that

[inland matrix]

Damn

[inland matrix]

If you're allowed a calculator then, idk

But if not then yes, sometimes

[grizzled tinsel]

ok

ill try to learn in my spare time i guess

[inland matrix]

Sure

[grizzled tinsel]

so y= 2x +1

which is correct

ok thanks

.close

[obtuse pebble]

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[marsh geyser]

That one u solve factoring it

[obtuse pebble]

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[dusky marten]

Hello, I need help understanding the dimensionality and reason for applying a transpose on lambda

[dusky marten]

this problem refers to the sensitivity of an adjoint system

in the following:

dj/dtheta has dimensions 1xP. dJ/dx has 1xN, df/dx is NxN and df/dtheta is NxP dimensional

so multiplying dJ/dx with (df/dx)^-1 yeilds a 1xN martix which can be multiplied with NxP to give 1xP which satisfies the equation

but taking the transpose of lambda gives dimensions Nx1 which cannot be multiplied with 1xP

lambda_T is defined as: -dJ/dx * (df/dx)^-1

<@&286206848099549185>

[elder crypt]

yep

[dusky marten]

my question is why do we transpose lambda

it changes the dimensionality meaning we cant multiply

[elder crypt]

which grade

[dusky marten]

what?

[dusky marten]

wdym grade?

[elder crypt]

from which country??

[dusky marten]

the topic is adjoint sensitivty

gradient based optimisation

<@&286206848099549185>

[obtuse pebble]

@dusky marten Has your question been resolved?

[dusky marten]

<@&286206848099549185>

[timid silo]

I will beat u up

[dusky marten]

bro open a new help channel

im using this one

read #❓how-to-get-help

<@&268886789983436800>

[dusky marten]

<@&286206848099549185> can some explain this please?

can someone explain to me why in adjoint sensitivity, we define lamdba_transpose instead of just lambda? i dont understand how the dimensions work? <@&286206848099549185>

[obtuse pebble]

@dusky marten Has your question been resolved?

[obtuse pebble]

@dusky marten Has your question been resolved?

[obtuse pebble]

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[honest forum]

Consider two solid spherical balls, one centered at $\left( 0, 0, \frac{21}{2} \right),$ with radius $6$, and the other centered at $(0,0,1)$ with radius $\frac{9}{2}.$ How many points $(x,y,z)$ with only integer coordinates are there in the intersection of the balls?

[warm shale]

Dork9399

[honest forum]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[honest forum]

I'm on step 2

I can visualize the two objects, but I don't know how to find the number of lattice points inside the intersection.

[obtuse pebble]

@honest forum Has your question been resolved?

[honest forum]

<@&286206848099549185>

[brazen viper]

@honest forum you can find the intersection of the two balls as a disc. Then once you have that disc, which is constrained to a plane z = a for some number a. If a is an integer, you just count the lattice points on that disc as if it were centered at the origin. If a is not an integer, then you have 0.

[honest forum]

Wouldn't the intersection be an object, not a disc?

[brazen viper]

what is the definition of object?

[honest forum]

Plotting it, I get this shape as the intersection

.close

[obtuse pebble]

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[brazen viper]

ah, my bad

you'll need to look at several circles. This is a bit complicated.

[obtuse pebble]

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[timid silo]

My question is if there is any site or place where i can learn about union set. And where it doesn't cost money cuz i am poor 🙂

[ruby path]

union set?

[timid silo]

this

[spring steeple]

Meant to say 'set union' I guess

[ruby path]

no i meant lik

[timid silo]

like your profile picture

[ruby path]

set theory in general?

for math you can always back on Khan Academy

[spring steeple]

https://youtu.be/xZELQc11ACY?feature=shared

[timid silo]

thank you very much

[timid silo]

thx, i like to use that

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[cosmic veldt]

A journal article reports that a sample of size $n=5$ was used as a basis for calculating a 95% confidence interval for the true mean natural frequency (Hz) of delaminated beams of a certain type. The resulting confidence interval was $(229.764, 233.504)$. You decide that a confidence level of $99%$ is more appropriate than the $95%$ level used. What are the limits of the $99%$ interval? Assume that the data collected follows a Normal distribution.

[warm shale]

cookie2

[obtuse pebble]

@cosmic veldt Has your question been resolved?

[cosmic veldt]

<@&286206848099549185>

[obtuse pebble]

@cosmic veldt Has your question been resolved?

[obtuse pebble]

@cosmic veldt Has your question been resolved?

[cosmic veldt]

.clsoe

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[real kelp]

Why do I get Sin (35) * 3 different from Sin (65) * 2 here?

[indigo night]

does this triangle exsist?

[real kelp]

Well, I made it

But it it satisfies the proportion rules

[indigo night]

fine

[real kelp]

a + b > c and all angles sum 180

Right?

[indigo night]

no

[real kelp]

Wait what?

[indigo night]

if you draw it

maybe the angles are not 35 and 65

use pencil and ruler

[real kelp]

Oh wait

Thats because when you have 3 given sides, then there's a unique solution for the three angles right?

[indigo night]

yes

[real kelp]

That makes sense cause I made up the sides and the angles

[real kelp]

Oh I see

Thxxx

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[feral harbor]

funnily #help-7｜zen1thxyz has the same problem

I would help you , but im very dumb. Sorry.

what?

[hexed gull]

.-.

[upbeat plinth]

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[spark horizon]

how do i do 10b

[obtuse pebble]

@spark horizon Has your question been resolved?

[spark horizon]

<@&286206848099549185>

[obsidian isle]

<@&286206848099549185>

[grand torrent]

!15mins

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[obtuse pebble]

@spark horizon Has your question been resolved?

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[outer needle]

What is your question bro

Just a tip to make getting help more likely

[cosmic veldt]

How do I show that S5 is generated by the set {(12)(23)(34)(45)}?

[zenith raft]

is there a typo in the problem...?

[jaunty oasis]

where are the commas in the set

[zenith raft]

should there be commas separating the transpositions

[jaunty oasis]

sully

[zenith raft]

no way that's supposed to be one element

S_5 is not cyclic

[jaunty oasis]

s5 is cyclic trust

[zenith raft]

i do not trust you

[jaunty oasis]

damn

[cosmic veldt]

Wait im also confused too

[jaunty oasis]

anyway yeah those are probably 4 elements?

[cosmic veldt]

r there supposed to b e commas lol

[zenith raft]

yes

[cosmic veldt]

this was the question given

no modifications

[zenith raft]

then it is poorly posed

[cosmic veldt]

I think then let's just say there's commas

then what?

i know that this thing is a theorem

but idk how to shuow it

[zenith raft]

are there any sets you know S_5 is generated by

other than the whole thing

[cosmic veldt]

that one

LOL

[zenith raft]

[cosmic veldt]

by theorem obviously

no intuition

Like "every sN can be written as a product of transpotition"

i think

[zenith raft]

yes great

[cosmic veldt]

n-1 transpositions*

[zenith raft]

so S_5 is generated by its transpositions

[zenith raft]

so you could just show you can make any transposition from these 4

[cosmic veldt]

u mean like 13, 14, 15, 23, 24, 25, 34, 35, 45?

[zenith raft]

think you're missing one but yea

(1 2)

[night fox]

cyclic permutations 👀

maybe the set is also permuted in some manner

and you want to prove that symmetric group under 5 is generated by the other permuted set.

[zenith raft]

[night fox]

like the transpositions are permuted as well

[zenith raft]

[night fox]

huh

[obtuse pebble]

@cosmic veldt Has your question been resolved?

[steep owl]

no, it’s permutation cycle notation

oh i see

[night fox]

yea

?

[steep owl]

yeah i think it’s a set of five different transpositions

lol

it’s basically saying “bubble sort terminates”

[night fox]

yea that works as well

[steep owl]

to prove this just use an algorithm (bubble sort) to get any element of S5

[night fox]

..

[steep owl]

yes i think so

[obtuse pebble]

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[hoary kindle]

how can i find the exact value of cosec (-60)

[olive dagger]

cosec(x) is an odd function

so cosex(-x) = -cosec(x)

[hoary kindle]

yep

[olive dagger]

now plug cosec(60)

csc(x) = 1/sin(x)

if u dont remember the values

[hoary kindle]

so it would be -1/sin(60)?

[olive dagger]

yes

[hoary kindle]

and then i use exact values for sin60

[olive dagger]

yup

[hoary kindle]

ohh alr

thanks

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[hexed gull]

xd

[obtuse pebble]

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[fresh lotus]

How do I factor when a doesn’t equal one

[median dome]

What?

Context?

[hexed gull]

um, idk, you let the penguin eat the 1?🐧

[fresh lotus]

Ohhhhhh I get it now

[fresh lotus]

I don’t really know how to give context…… Like my pre algebra class has been factoring variables and numbers and now we are gonna do things like positive c, negative c, and when a doesn’t equal one.

But I don’t really know what that means because when we are factoring it’s not like we are trying to figure out what the variable is anyways, so why do we need to learn cases when a doesn’t equal one

[median dome]

What is a…

The leading coefficient?

[obsidian isle]

yeah so there's multiple ways you can go about this

the most common method is to split the linear term

[worn yoke]

are you asking how to factor quadratics in the form
[ ax^2 + bx + c ]
where $a\ne 1$...? you should tell us what you mean, we don't know what `$a$' is automatically

[warm shale]

cloud

[fresh lotus]

I don’t even know what quadratics are….😢

Oh I searched it up

[edgy needle]

non monic?

[fresh lotus]

I think

I can just ask my teacher on Monday

My quiz is on Thursday so I got time

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[dry creek]

does anybody know where i went wrong

[radiant osprey]

This image is illegible

[zenith raft]

darn i wanted to make a joke about how it was just legible enough for me to make out that it says final exam at the top

but it also says 2023

[dry creek]

ok sorry about that, is this better

[hexed gull]

hm haven't seen the definition of centroid, is it the point for which the integral w.r.t. to x & are equal on both sides?

yeah makes the most sense to me

well x is fixed due to symmetry

and I presume you determined y by setting the two integrals w.r.t y equal?

or have you been given a specific other formula for centroid coordintaes @dry creek

[hexed gull]

hm I don't want to write a proof for it, but I presume that definition is synonymous to the integrals being equal in x & y

[dry creek]

wdym being equal in x and y exactly?

[hexed gull]

such that A1 = A2

and

A1 = A2 here

which would be applicable

[dry creek]

um

probably

[hexed gull]

for an arbitrary shape

[dry creek]

our prof didnt really go over too much what it really means

but i do know the formula is guarenteed to work

[hexed gull]

but centroid is just weight center right

so it makes sense for the integrals to be equal

with respect to any orthogonal basis

it'd be the same in 3D

or 4D

[dry creek]

we havent learned orthogonal stuff lol he just gave us the formula

[hexed gull]

kk, but it's clear that xdash = 0 right

[dry creek]

which xdash?

vertical line in first image?

[dry creek]

uh

[hexed gull]

the x-coordinate of the centroid

= xdash

[dry creek]

oh its asking for y coord here

[hexed gull]

yea but I mean it's clear why xdash = 0 right

[dry creek]

ok

[hexed gull]

well because the shape is symmetrical

y = sqrt(4-x²) is symmetrical to the y-axis

and y = 1 as well

[dry creek]

ok

[hexed gull]

thereby the enclosed shape is symmetrical as well

and the centroid must lie on the y-axis

now, if you imagine the centroid as "weight center"

it also makes sense that ydash must be in between here:

on this red line

[dry creek]

ok

[hexed gull]

since the weight center of that half disk must lie on the disk

imagine balancing that shape on one finger

if you want to stay it on your finger, then you need to place it somewhere on that red line

otherwise it'll tip over

nywys back to the task lol

ok so how far did you get in evaluating ydash

with the formula you provided:

[dry creek]

i evaluated it fully but its wrong

[hexed gull]

hm k do you happen to have the solution

I'd just like to briefly check whether my former thesis is true

[hexed gull]

aw no

sadly not

hm then centroid prob isn't the same as weight center :/

[hexed gull]

k I'll check your steps now

[dry creek]

ok

if you need me to make any of the steps that are maybe blurry clearer lmk

[hexed gull]

1.3835 would be the weight center btw

which is suspiciously close

[hexed gull]

where is your final value for the area?

[dry creek]

its -6pi/3-sin(8pi/3)-3sqrt3/2

[hexed gull]

but that's negative

[dry creek]

yeah lol

something went wrong but i cant figure it out

[hexed gull]

the area is just 2 * int[1 to 2]f(x)dx where f(x) is the circle function

[dry creek]

i made it from -sqrt3 to sqrt3 along the x axis instead of y

[hexed gull]

yea works as well

[dry creek]

which is harder but should still work

[hexed gull]

hm the substitution for theta seems right but I'm not sure what for

[dry creek]

because i didnt know what to do with the sqrt

[hexed gull]

but you can just evaluate it

directly

without a calculator

[dry creek]

how would i do the sqrt part without trig sub?

[hexed gull]

hm I may have done expansion (if they want the solution just as estimate rather than exact value

but exact value trig sub yeah k

[dry creek]

yeah they want exact value

[hexed gull]

hm don't yet see it, am at the values

am confused where the issue occurs

[dry creek]

me too lol

the fact that i got sin(8pi/3) which i dont think is a value of the unit circle is werid since usually these weird values dont show up for these problems

and the negative area lol

[hexed gull]

8pi/3 = pi/3 for sin

so sin(8pi/3) = sqrt(3)/2

[dry creek]

oh ok

[hexed gull]

I think it has to do with the periodic nature of sin/cos?

that you can switch with - & 2pi in some places and get the right value

but I'm not sure where that flaw occurs

[dry creek]

yeah the answer key just like directly solves for it

[hexed gull]

wait why is it - - here

ah nvm

yeah getting 4pi/3-sqrt(3) here shall be the intended

don't see it sry

[dry creek]

wait do you mean the answer should be 4pi/3 - sqrt3?

[dry creek]

oh yeah

[hexed gull]

good luck with finding it, gtg in a sec

[dry creek]

alr

[obtuse pebble]

@dry creek Has your question been resolved?

[obtuse pebble]

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[open kelp]

How should I start with this?
kinda stuck on what to do.

[brisk matrix]

what are the conditions an ideal must satisfy?

show that the intersection satisfies those conditions

my sir can probably help you with this @fathom flicker

he has a doubt about ideals

[zenith raft]

@fathom flicker

[fathom flicker]

Let me ask my sir @gilded needle .. I have a doubt about what a "left" ideal is

[open kelp]

how to differentiate with left and right ideals?

[fathom flicker]

is it such that for all i in I, that for all a in R,
ai in I

yes

your ring is non-commutative

it is not necessarily the same that ri=ir

does this clear your doubt?

[open kelp]

if i were to solve it, i would use A⋂B?

[brisk matrix]

wdym

[gilded needle]

the intersection of arbitrarily many left ideals is surely a left ideal, if my sir is not very mistaken

[open kelp]

Let A and B be ideals of R. Then ∃ 0 ∈ R such that 0 ∈ A and 0 ∈ B. Thus 0 ∈ A ⋂ B. Hence A⋂B is nonempty.

something like that?

[brisk matrix]

you dont need a quantifier

[open kelp]

or do i specify that A and B are left ideals...

[brisk matrix]

just 0 is in R, so 0 is in A and B

[brisk matrix]

you should do this as well

[open kelp]

so something like this?:
Let A and B be left ideals of R. Then 0 ∈ A and 0 ∈ B. Thus 0 ∈ A ⋂ B. Hence A⋂B is nonempty.

[brisk matrix]

yes, you might want to mention why 0 is in A and B, but otherwise is fine

next you'd show that A n B is closed under addition

and that it is closed under additive inverses, and finally under left multiplication from R

[open kelp]

Let x,y ∈ A⋂B. ... here's where I'm kinda stuck

[brisk matrix]

well if x,y are in A n B

then they are in A

what do you know about x + y, if x,y are in A

[open kelp]

it is also in B?

[brisk matrix]

no

well yes, but that isn't what i was asking about

you have x,y in AnB

that means that x,y are in A

look at the second condition for an ideal

[open kelp]

so x+y in AnB?

[brisk matrix]

how did you get to that

[open kelp]

Let x,y be elements of AnB such that x+y are in A and in B. thus x+y is in AnB? not sure... i just went in directly

[brisk matrix]

follow along with me, and answer the questions i ask

let x,y be in A

assuming A is a left ideal of R

is x + y in A?

[open kelp]

yes.

[brisk matrix]

why

you are correct

but why

[open kelp]

since it is an ideal of R

[brisk matrix]

ok yes

it is the second condition

so you would word your proof as

let x,y be in AnB.

• then x,y are in A, so x + y is in A
• then x,y are in B, so x + y is in B
  so x + y is in A and B
  so x + y is in AnB

[open kelp]

its a much clearer version of this

i see it now

Let x,y in AnB. Then x,y in A and x,y in B. thus there exist -x,-y in A and -x,-y in B?

[brisk matrix]

note that you don't need both x and y now

you can just let x be in AnB

otherwise it looks fine

[open kelp]

Let x in AnB. Then x in A and x in B. thus there exist -x in A and -x, in B. We have x+(-x) = 0 in AnB

[brisk matrix]

yeah looks fine

[open kelp]

Let x,y in AnB such that xy in AnB. Let z in AnB such that z(xy). Thus, z(xy) = (zx)y in AnB.

Is this good enough?

[brisk matrix]

no

you are trying to show that if x in AnB, and r in R, then rx is in AnB

start by letting x be in AnB and r in R

[open kelp]

oh my bad...

[brisk matrix]

since x in A, ra in A, and since x in B, ...

[brisk matrix]

np

[brisk matrix]

it is very similar to the other ones we did

you just break it down to the case in A and B, and show that ra is in A and B, and so in AnB

[open kelp]

Let x,y in AnB, then x,y in A implies that xy in A and x,y in B implies that xy in B.

right?

[brisk matrix]

you dont need x and y

you need an x in AnB, and an r in R

[open kelp]

Let x,y in AnB and r in R. Then x in A implies that rx in A and x in B implies that rx in B.

[brisk matrix]

yes

get rid of x,y

just x

[open kelp]

Let x in AnB and r in R. Then x in A implies that rx in A and x in B implies that rx in B. Thus rx in AnB.
Since A and B are noncommutative, then AnB is a left ideal of R

[brisk matrix]

Since A and B are noncommutative, then AnB is a left ideal of R
this doesnt really make sense and i would drop it

i would just say that AnB satisfies all conditions to be a left ideal

[open kelp]

its kinda the conclusion

[open kelp]

okay^2

thank you so much

.close

[obtuse pebble]

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[obtuse pebble]

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[strong forum]

can someone explain the difference in approaches for part c of each question, why did we treat X as exact same for one but not for other?

[remote skiff]

I don't understand what you are asking

[obtuse pebble]

@strong forum Has your question been resolved?

[strong forum]

For 1C on the first very first image, the solution is Var(V)=4Var(Y) + 4Var(U) instead of using V=10X-8

[remote skiff]

There is nothing particular about either problem that makes one approach more special than the other. That one is missing the justification that Cov(Y,U) = 0.

[strong forum]

if I try doing

V=10X-8

I get Var(V) = 10^2 * 16

which isn't 832

[remote skiff]

Like I already said, "That one is missing the justification that Cov(Y,U) = 0."

[strong forum]

what does Cov(Y,U) mean sorry?

that they are independent?

[remote skiff]

The covariance of Y and U

[strong forum]

Alright, thjank u!

.close

[obtuse pebble]

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[strong forum]

.reopen

[obtuse pebble]

✅

[strong forum]

Sorry just one more thing, why do both methods not result in the same answer?

If independent we do that method if not we do the one I suggested?

[remote skiff]

I said nothing about independence. If the answers aren't correct it's probably because they did the calculation wrong and assumed Cov(Y,U) = 0 when it's not.

you should do the calculation yourself as a matter of practice

[strong forum]

so value of covariance effects whether you use Var(10X-8) or 4Var(Y)+4Var(U) using the Var(aX+bY)=a^2Var(X) + b^2Var(Y) property?

because I did the calculations and both of those methods lead to different answers

I also never heard of covariance before and assumed cov(y,u)=0 meant y and u are independent since thats when var(aX+bY) thing works

at least i was taught

[remote skiff]

No, the property is Var(aX+bY) = a^2 Var(X) + b^2Var(Y) +2abCov(X,Y). They omitted that term, so it is either zero or they are wrong.

zero covariance does not mean independence

[strong forum]

Ohh

that makes sense

thank you!

[obtuse pebble]

@strong forum Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[silk monolith]

how do I solve this?

[olive dagger]

express tan4x in terms of sine and cosine

[silk monolith]

should i use tan(2θ-2θ) or smth?

[fierce elbow]

maybe tan [2(2θ)]

[silk monolith]

or tan4θ=(sin4θ)/(cos4θ)

[silk monolith]

lemme try

[fierce elbow]

wait

is there a formula for than 2θ

[silk monolith]

actually i dont think so for tan

cos and sin only

[fierce elbow]

there is

[timid silo]

square the first equation

[olive dagger]

yea

[olive dagger]

yes there is

[timid silo]

you get 1 - sin2x = k^2

now its easier

[olive dagger]

oh yeah

fr

[silk monolith]

where did that come from bro?

oh wait you sqaured both sides

[timid silo]

ye

[silk monolith]

yeah but what do I do with that?

[timid silo]

change tan4x to 2tan2x/1-tan^2(2x)

you know sin2x and cos2x

[silk monolith]

ohhh damn

aight bro gimme a sec lemme calculate it

[obtuse pebble]

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[obtuse pebble]

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[pine badger]

is this correct? did i make a mistake somewhere?

and can it be simplified further?

[last pilot]

firstly

u' is not 12x cos x

it needs to be simplified further

u = fg
f = 6x^2
g = sin x

use product rule

[obtuse pebble]

@pine badger Has your question been resolved?

[pine badger]

Like this?

[pine badger]

<@&286206848099549185>

[pine badger]

<@&286206848099549185>

@last pilot

[obtuse pebble]

@pine badger Has your question been resolved?

[obtuse pebble]

@pine badger Has your question been resolved?

[obtuse pebble]

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[haughty comet]

Hi, I'm trying to minimize/simplify a boolean expression I got from a karnaugh map to see if i can reduce it any further.
This is my working so far and I can't tell if im just stuck or it can't be simplified further.

[haughty comet]

<@&286206848099549185>

[obtuse pebble]

@haughty comet Has your question been resolved?

[obtuse pebble]

@haughty comet Has your question been resolved?

[obtuse pebble]

@haughty comet Has your question been resolved?

[haughty comet]

.close

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[wooden cipher]

!filetype

[obtuse pebble]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[timid silo]

.close

[obtuse pebble]

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[polar fossil]

@timid silo in the future, please do not hop around to other channels and advertise your current one or interrupt what's happening over there

[obtuse pebble]

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[worldly lagoon]

How are either of these wrong

It makes 0 sense to me

[surreal forge]

maybe they want you to fully simplify

[worldly lagoon]

I don't really understand what it would entail to "fully simplify" here

square the 5tan(theta)

?

[surreal forge]

r u familiar with trignometric pythagorean identity

[worldly lagoon]

a^2+b^2?

[surreal forge]

(sinx)^2 + (cosx)^2 = 1

[worldly lagoon]

yeah that I know

[surreal forge]

divide both side by (cosx)^2

u will see how to simplify ur expression above

[worldly lagoon]

I'm sorry I don't understand

both sides of what

[surreal forge]

of this equation

[worldly lagoon]

ok sure

how does that help

[surreal forge]

something seems very incorrect here

[worldly lagoon]

ah wait 1 sec

but that's just this identity

mmmm

I see what you mean

make it just sec

but its not tan^2

I mean 5sec(theta)

is all that comes to mind

because 25+25tan^(theta)

take the 5 out

ok I got it

.close

[obtuse pebble]

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[obtuse pebble]

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[modest minnow]

what can i cancel out in this equation. im confused by the plus sign in the numerator
n(n + 2) + (n + 1) / (n + 1 )(n +2)

[fluid plume]

hello

[modest minnow]

hello

[fluid plume]

is this what you got?

$$\frac{n(n+2)+(n+1)}{(n+1)(n+2)}$$

[warm shale]

dp

[modest minnow]

yea thats right

[fluid plume]

ok

You're right, we can't cancel anything here

but, maybe if we do the sum in the numerator we get something we can cancel

try it, and see if you figure out anything

if you need help, tell me

[modest minnow]

ok

wait but would i need to do the multiplication of n(n+2) first?

[fluid plume]

yes, we would do n(n+2) first, and then add (n+1)

PEMDAS